Equal masses of `3` liquids `A,B` and `C` have temperatures `10^@C, 25^@C` and `40^@C` respectively. If `A` and `B`are mixed, the mixture has a temperature of `15^@C`. If `B` and `C` are mixed, the mixture has a temperature of `30^@ C`. If `A` & `C` are mixed the temperature of the mixture is

Options

A. `16^@ C`
B. `35^@C`
C. `20^@ C`
D. `25^@C`

Explanation

To solve the problem step by step, we will use the principle of calorimetry, which states that the heat lost by the hotter substance will be equal to the heat gained by the cooler substance when they are mixed.

### Step 1: Define the variables and initial conditions
Let the masses of liquids A, B, and C be equal, denoted as \( M \). The temperatures of the liquids are:
- \( T_A = 10^\circ C \) (for liquid A)
- \( T_B = 25^\circ C \) (for liquid B)
- \( T_C = 40^\circ C \) (for liquid C)

### Step 2: Analyze the mixing of A and B
When liquids A and B are mixed, the final temperature \( T_{AB} \) is given as \( 15^\circ C \). We can set up the heat balance equation:
\[ M \cdot S_A \cdot (T_{AB} - T_A) = M \cdot S_B \cdot (T_B - T_{AB}) \]
Substituting the known values:
\[ M \cdot S_A \cdot (15 - 10) = M \cdot S_B \cdot (25 - 15) \]
This simplifies to:
\[ M \cdot S_A \cdot 5 = M \cdot S_B \cdot 10 \]
Canceling \( M \) from both sides (since they are equal) gives:
\[ 5 S_A = 10 S_B \implies S_A = 2 S_B \quad \text{(Equation 1)} \]

### Step 3: Analyze the mixing of B and C
When liquids B and C are mixed, the final temperature \( T_{BC} \) is given as \( 30^\circ C \). The heat balance equation is:
\[ M \cdot S_B \cdot (T_{BC} - T_B) = M \cdot S_C \cdot (T_C - T_{BC}) \]
Substituting the known values:
\[ M \cdot S_B \cdot (30 - 25) = M \cdot S_C \cdot (40 - 30) \]
This simplifies to:
\[ M \cdot S_B \cdot 5 = M \cdot S_C \cdot 10 \]
Canceling \( M \) gives:
\[ 5 S_B = 10 S_C \implies S_B = 2 S_C \quad \text{(Equation 2)} \]

### Step 4: Relate the specific heats
From Equation 1, we have \( S_A = 2 S_B \). Substituting Equation 2 into this gives:
\[ S_A = 2(2 S_C) = 4 S_C \quad \text{(Equation 3)} \]

### Step 5: Analyze the mixing of A and C
Now we need to find the final temperature \( T \) when A and C are mixed. The heat balance equation is:
\[ M \cdot S_A \cdot (T - T_A) = M \cdot S_C \cdot (T_C - T) \]
Substituting the known values:
\[ M \cdot S_A \cdot (T - 10) = M \cdot S_C \cdot (40 - T) \]
Canceling \( M \) gives:
\[ S_A (T - 10) = S_C (40 - T) \]
Substituting \( S_A = 4 S_C \) into the equation:
\[ 4 S_C (T - 10) = S_C (40 - T) \]
Dividing both sides by \( S_C \) (assuming \( S_C \neq 0 \)):
\[ 4 (T - 10) = 40 - T \]
Expanding and rearranging gives:
\[ 4T - 40 = 40 - T \]
\[ 4T + T = 80 \]
\[ 5T = 80 \implies T = 16^\circ C \]

### Final Answer
The temperature of the mixture when liquids A and C are mixed is \( 16^\circ C \).

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